[TEST] 小升初题目详解

问题

求解:

0+ess5 ds2++et22 dt0+sint2 dt(n=0(1)n2n+10+sinxx dx+n=1arctan2n2limt0+20202020tcosxx2+t2 dx)limn{[(01xn11+x dx)n12]n2}\frac{\frac{\displaystyle {\int_{0}^{+\infty} e^{-s} s^{5} \mathrm{~d} s}}{2}+\frac{\displaystyle \int_{-\infty}^{+\infty} e^{-\frac{t^{2}}{2}} \mathrm{~d} t}{\displaystyle \int_{0}^{+\infty} \sin t^{2} \mathrm{~d} t}\left(\frac{\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1}}{\displaystyle\int_{0}^{+\infty} \frac{\sin x}{x} \mathrm{~d} x}+\frac{\displaystyle\sum_{n=1}^{\infty} \arctan \frac{2}{n^{2}}}{\displaystyle\lim _{t \rightarrow 0^{+}} \displaystyle \int_{-2020}^{2020} \frac{t \cos x}{x^{2}+t^{2}} \mathrm{~d} x}\right)}{\displaystyle\lim _{n \rightarrow \infty}\left\{\left[\left(\displaystyle\int_{0}^{1} \frac{x^{n-1}}{1+x} \mathrm{~d} x\right) n-\frac{1}{2}\right] \frac{n}{2}\right\}}

Part 1. Gamma 函数部分

注意到 0+ess5ds=Γ(6)=5!=120\displaystyle\int_{0}^{+\infty}e^{-s}s^5\text{d}s=\Gamma(6)=5!=120.

Gamma 函数 Γ()\Gamma(\cdot) 又称为第二类欧拉积分, 定义如下:

Γ(s):=0exxs1dx\boxed{\Gamma(s):=\int_{0}^{\infty}e^{-x}x^{s-1}\text{d}x}

尝试分部积分得到:

\begin{align}{\color{Red} {\Gamma(s) }} & = \int_{0}^{\infty}e^{-x}x^{s-1}\text{d}x = -\left.e^{-x} x^{s-1}\right|_{0} ^{\infty}+(s-1) \int_{0}^{\infty}e^{-x} x^{s-2} \mathrm{~d} x\\&=(s-1) \int_{0}^{\infty} e^{-x} x^{(s-1)-1}\mathrm{~d} x={\color{Red} {(s-1)\Gamma(s-1)}} \end{align}

可以发现 Γ\Gamma 函数的递推性质. 又由定义知

Γ(1)=0+exdx=1\Gamma(1)=\int_{0}^{+\infty}e^{-x}\text{d}x=1

所以 sZ+\forall s\in \mathbf{Z}^+, 有 Γ(s+1)=s!\color{red}{\Gamma(s+1)=s!}.

Γ\Gamma 函数还可以这样定义:

Γ(x)=1xn=1(1+1n)x1+xn\Gamma(x)=\frac{1}{x}\prod_{n=1}^{\infty}\frac{\left ( 1+\frac{1}{n}\right )^x }{1+\frac{x}{n} }

这个定义和含参积分所给出的定义是可以互相导出的(过程待补充). 利用这个定义, 结合正弦函数的无穷乘积展开

sinx=xk=1(1x2k2π2)\sin x=x\prod_{k=1}^{\infty }\left ( 1-\frac{x^2}{k^2\pi ^2} \right )

还可以推导出所谓余元公式(过程待补充):

Γ(x)Γ(1x)=πsinπxx(0,1)\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x}\qquad x\in(0,1)

另外, 第一类欧拉积分(Beta 函数)

B(p,q)=01xp1(1x)q1dx\boxed{B(p,q)=\int_{0}^{1}x^{p-1}(1-x)^{q-1}\text{d}x}

与第二类欧拉积分之间有这样的关系(证明待补充):

B(p,q)=B(q,p)=Γ(p)Γ(q)Γ(p+q)B(p,q)=B(q,p)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}

最后, 我们还有所谓 Legendre 公式(证明待补充):

Γ(s)Γ(s+12)=π22s1Γ(2s)\Gamma(s)\Gamma(s+\frac{1}{2})=\frac{\sqrt{\pi}}{2^{2s-1}}\Gamma(2s)

Part 2. Gauss 积分部分

+et22dt=2π\displaystyle\int_{-\infty}^{+\infty}e^{-\frac{t^2}{2}}\text{d}t=\sqrt{2\pi}

其实是高斯积分的变体. 高斯积分指的是

+ex2dx=π\boxed{\int_{-\infty}^{+\infty}e^{-x^2}\text{d}x=\sqrt{\pi}}

  • 解法一: 利用二重积分.

    I=+ex2dxI=\int_{-\infty}^{+\infty}e^{-x^2}\text{d}x

    I2=(+ex2dx)2=+ex2dx+ey2dy=++e(x2+y2)dxdyI^2=\left ( \int_{-\infty}^{+\infty}e^{-x^2}\text{d}x \right )^2=\int_{-\infty}^{+\infty}e^{-x^2}\text{d}x\int_{-\infty}^{+\infty}e^{-y^2}\text{d}y=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)}\text{d}x{d}y

    利用极坐标变换转化积分

    I2=++e(x2+y2)dxdy=02π0er2rdr=02πdθ0er2rdrI^2=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)}\text{d}x{d}y=\int_{0}^{2\pi }\int_{0}^{\infty}e^{-r^2}r\text{d}r =\int_{0}^{2\pi }\text{d}\theta\int_{0}^{\infty}e^{-r^2}r\text{d}r\\

    最右端第二个积分作换元 u=r2u=-r^2 就容易解得

    0er2rdr=12\int_{0}^{\infty}e^{-r^2}r\text{d}r=\frac{1}{2}

    最终得到

    I2=2π12=πI^2=2\pi\cdot\frac{1}{2}=\pi

    由于 ex20e^{-x^2}\geq 0, 所以 I0I\geq 0, 故

    I=+ex2dx=π\color{red}{I=\int_{-\infty}^{+\infty}e^{-x^2}\text{d}x=\sqrt{\pi}}

  • 解法二: 利用 Gamma 函数
    t2=αt^2=\alpha, 则

    +et2dt=+12eαα12dα=0+eαα121dα=Γ(12)\color{red}{\int_{-\infty}^{+\infty }e^{-t^2}\text{d}t}=\int_{-\infty}^{+\infty }\frac{1}{2}e^{-\alpha}\alpha^{-\frac{1}{2}}\text{d}\alpha=\int_{0}^{+\infty }e^{-\alpha}\alpha ^{\frac{1}{2}-1}\text{d}\alpha=\color{red}{\Gamma\left(\frac{1}{2}\right)} \\

    由余元公式可得

    [Γ(12)]2=Γ(12)Γ(112)=πsinπ2=π\left [ \Gamma\left ( \frac{1}{2} \right ) \right ]^2= \Gamma\left ( \frac{1}{2} \right )\Gamma\left(1-\frac{1}{2} \right)= \frac{\pi }{\sin\frac{\pi}{2} }=\pi \\

    故有

    +et2dt=Γ(12)=π\color{red}{\int_{-\infty}^{+\infty }e^{-t^2}\text{d}t}={\Gamma\left(\frac{1}{2}\right)}=\color{red}{\sqrt{\pi}}

  • 解法三: 利用 Laplace 变换

    f(t)=0etx2dxf(t)=\int_{0}^{\infty }e^{-tx^2}\text{d}x\\

    对其作 Laplace 变换:

    L[f(t)]=L(0etx2dx)=00etx2estdtdx=0L(etx2)dx\mathcal{L} [f(t)]=\mathcal{L}\left ( \int_{0}^{\infty }e^{-tx^2}\text{d}x \right ) =\int_{0}^{\infty }\int_{0}^{\infty }e^{-tx^2}e^{-st}\text{d}t\text{d}x=\int_{0}^{\infty }\mathcal{L} \left ( e^{-tx^2} \right )\text{d}x\\

    运用常用 Laplace 变换公式

    L(eat)=1sa\mathcal{L} \left ( e^{at} \right )=\frac{1}{s-a} \\

    可得

    L[f(t)]=0L(etx2)dx=0dxs+x2=π2s\mathcal{L} [f(t)]=\int_{0}^{\infty }\mathcal{L} \left ( e^{-tx^2} \right )\text{d}x=\int_{0}^{\infty }\frac{\text{d}x}{s+x^2}=\frac{\pi }{2\sqrt{s}}

    再对两边作 Laplace 逆变换可得

    f(t)=0etx2dx=π2tf(t)=\int_{0}^{\infty } e^{-tx^2} \text{d}x=\frac{\sqrt \pi }{2\sqrt{t}}

    在上式中令 t=1t=1 即得

    +ex2dx=20+ex2dx=2π2=π\color{red}{\int_{-\infty}^{+\infty}e^{-x^2}\text{d}x}=2\int_{0}^{+\infty}e^{-x^2}\text{d}x=2\cdot\frac{\sqrt{\pi}}{2}=\color{red}{\sqrt{\pi}}

    还有一些形式类似的积分:

    +eax2+bx+cdx=πaeb24a+c\int_{-\infty }^{+\infty}e^{-a x^{2}+bx+c} \mathrm{d} x =\sqrt{\frac{\pi }{a} }e^{\frac{b^2}{4a}+c }\\

    0x2nex2a2 dx=πa2n+1(2n1)!!2n+1\int_{0}^{\infty} x^{2 n} e^{-\frac{x^{2}}{a^{2}}} \mathrm{~d} x =\sqrt{\pi} \frac{a^{2 n+1}(2 n-1) ! !}{2^{n+1}} \\

    0x2n+1ex2a2 dx=n!2a2n+2\int_{0}^{\infty} x^{2 n+1} e^{-\frac{x^{2}}{a^{2}}} \mathrm{~d} x =\frac{n !}{2} a^{2 n+2} \\

    0x2neax2 dx=(2n1)!!an2n+1πa\int_{0}^{\infty} x^{2 n} e^{-a x^{2}} \mathrm{~d} x =\frac{(2 n-1) ! !}{a^{n} 2^{n+1}} \sqrt{\frac{\pi}{a}} \\

    0x2n+1eax2 dx=n!2an+1\int_{0}^{\infty} x^{2 n+1} e^{-a x^{2}} \mathrm{~d} x =\frac{n !}{2 a^{n+1}} \\

    0xneax2 dx=Γ(n+12)2an12\int_{0}^{\infty} x^{n} e^{-a x^{2}} \mathrm{~d} x =\frac{\Gamma\left(\frac{n+1}{2}\right)}{2 a^{\frac{n-1}{2}}}\\

Part 3. Fresnel 积分部分

0+sint2dt=π8\displaystyle\int_{0}^{+\infty}\sin t^2\text{d}t = \sqrt{\frac{\pi }{8}}

同样有很多种解法, 这里先写一种利用高斯积分的: 作换元 t2=αt^2=\alpha, 则积分变为

0+sint2dt=120+sinαadα\int_{0}^{+\infty}\sin t^2\text{d}t=\frac{1}{2}\int_{0}^{+\infty}\frac{\sin\alpha}{\sqrt{a}}\text{d}\alpha

由高斯积分的结果

0+ex2dx=π2\int_{0}^{+\infty}e^{-x^2}\text{d}x=\frac{\sqrt{\pi}}{2}

x=αux=\sqrt{\alpha}u

α0+eαu2du=π2\sqrt{\alpha }\int_{0}^{+\infty} e^{-\alpha u^2}\text{d}u=\frac{\sqrt{\pi }}{2}

所以

1α=2π0+eαu2du\frac{1}{\sqrt{\alpha }}=\frac{2}{\sqrt{\pi }} \int_{0}^{+\infty} e^{-\alpha u^2}\text{d}u

据此改写原题积分

\begin{align}\int_{0}^{+\infty}\sin t^2\text{d}t & = \frac{1}{2}\int_{0}^{+\infty}\frac{\sin\alpha}{\sqrt{\alpha}}\text{d}\alpha\\&=\frac{1}{2}\cdot \frac{2}{\sqrt{\pi }} \int_{0}^{+\infty}\sin t\int_{0}^{+\infty}e^{-tu^2}\text{d}u\text{d}t\\&=\frac{1}{\sqrt{\pi }}\int_{0}^{+\infty}\left ( \int_{0}^{+\infty}\sin t\cdot e^{-tu^2}\text{d}t \right ) \text{d}u\end{align}

利用正弦函数的 Laplace 变换

0eaxsinxdx=11+a2\int_{0}^{\infty} e^{-ax}\sin x\text{d}x=\frac{1}{1+a^2}

可得

0+sint2dt=1π0+(0+sintetu2dt)du=1π0+11+u4du=π8\color{red}{\int_{0}^{+\infty}\sin t^2\text{d}t} = \frac{1}{\sqrt{\pi }}\int_{0}^{+\infty}\left ( \int_{0}^{+\infty}\sin t\cdot e^{-tu^2}\text{d}t \right ) \text{d}u = \frac{1}{\sqrt{\pi }}\int_{0}^{+\infty}\frac{1}{1+u^4} \text{d}u =\color{red}{ \sqrt{\frac{\pi }{8}}}

我们还可以用留数定理来解这个积分: 由 Euler 公式容易得到

eix2=cosx2+isinx2e^{{ix}^2}=\cos x^2+i\sin x^2

两边同时积分得到

0eix2dx=0cosx2dx+i0sinx2dx\int_{0}^{\infty }e^{ix^2}\text{d}x= \int_{0}^{\infty }\cos x^2\text{d}x+i\int_{0}^{\infty }\sin x^2\text{d}x\\

考虑围道积分 Ceiz2dz\displaystyle\oint_{C}e^{iz^2}\text{d}z, 选取扇形围道如下图:

积分路径按逆时针方向: 从 O 点出发行进至 R 处, 经圆弧 C_R 再返回 O 点. 由于围道内无奇点, 所以

Ceiz2dz=0Reix2dx+CReiz2dz+R0eiz2dz=0\oint_{C}e^{iz^2}\text{d}z=\int_{0}^{R}e^{ix^2}\text{d}x+\int _{C_R}e^{iz^2}\text{d}z+\int_{R}^{0}e^{iz^2}\text{d}z=0

limR0Reix2dx=0eix2dx=0cosx2dx+i0sinx2dx\lim_{R \to \infty} \int_{0}^{R} e^{ix^2}\text{d}x=\int_{0}^{\infty } e^{ix^2}\text{d}x=\int_{0}^{\infty } \cos x^2\text{d}x +i\int_{0}^{\infty }\sin x^2\text{d}x

\begin{align}\lim_{R \to \infty} \int_{R}^{0} e^{iz^2}\text{d}z & = \lim_{R \to \infty} \int_{R}^{0} e^{i\left ( Re^{\frac{i\pi}{4}} \right ) ^2}\text{d}\left ( Re^{\frac{i\pi}{4}}\right ) = \lim_{R \to \infty} \int_{0}^{R}e^{i\cdot iR^2}\cdot e^{\frac{i\pi}{4}}\text{d}R\\& = -e^{\frac{i\pi}{4}}\int_{0}^{\infty }e^{-R^2}\text{d}R=-\left ( \cos\frac{\pi }{4}+i\sin \frac{\pi }{4}\right )\frac{\sqrt{\pi }}{2} \\&=-\frac{\sqrt{\pi }}{2\sqrt{2}}(1+i) =\sqrt{\frac{\pi }{8} } +i\sqrt{\frac{\pi }{8} } \end{align}

CReiz2dz=0(Jordan’s Lemma)\int _{C_R}e^{iz^2}\text{d}z=0\qquad (\text{Jordan's Lemma})

所以

0cosx2dx+i0sinx2dx=π8+iπ8\int_{0}^{\infty }\cos x^2\text{d}x +i\int_{0}^{\infty }\sin x^2\text{d}x=\sqrt{\frac{\pi }{8} } +i\sqrt{\frac{\pi }{8} }

在上式中取虚部即得

0+sint2dt=π8\color{red}{\int_{0}^{+\infty}\sin t^2\text{d}t = \sqrt{\frac{\pi }{8}}}

同时告诉我们

0+cost2dt=π8\int_{0}^{+\infty}\cos t^2\text{d}t = \sqrt{\frac{\pi }{8}}

另外, 还有

0xsint2dt=n=0(1)nx4n+3(2n+1)!(4n+3)\int_{0}^{x}\sin t^2\text{d}t =\sum_{n=0}^{\infty }(-1)^n\frac{x^{4n+3}}{(2n+1)!(4n+3)} \\

0xcost2dt=n=0(1)nx4n+1(2n)!(4n+1)\int_{0}^{x}\cos t^2\text{d}t =\sum_{n=0}^{\infty }(-1)^n\frac{x^{4n+1}}{(2n)!(4n+1)} \\

对任意 xx 均收敛.

Part 4. 反正切函数的 Taylor 展开

注意到

arctanx=n=0(1)n2n+1x2n+1\arctan x=\sum_{n=0}^{\infty }\frac{(-1)^n}{2n+1}x^{2n+1}

所以 n=0(1)n2n+1=arctan1=π4\color{red}{\displaystyle\sum_{n=0}^{\infty }\frac{(-1)^n}{2n+1}}=\arctan 1=\color{red}{\frac{\pi}{4}}.

Part 5. Dirichlet 积分部分

注意到 Dirichlet 积分 0+sinxxdx=π2\displaystyle\int_{0}^{+\infty}\frac{\sin x}{x}\text{d}x=\frac{\pi}{2}.

  • 解法一: 利用 Lobachevsky 积分法
    Lobachevsky 积分法阐述如下: 若函数 f(x)f(x)[0,+)[0,+\infty) 上有

    f(πx)=f(x),f(x+π)=f(x)f(\pi -x)=f(x),\qquad f(x+\pi)=f(x)

    0+f(x)sinxxdx=0π2f(x)dx\int_{0}^{+\infty}f(x)\frac{\sin x}{x}\text{d}x=\int_{0}^{\frac{\pi}{2}}f(x)\text{d}x

    事实上, 只需在上式中令 f(x)=1f(x)=1 即可得到

    0+sinxxdx=π2\int_{0}^{+\infty}\frac{\sin x}{x}\text{d}x=\frac{\pi}{2}

    我们附上 Lobachevsky 积分法的证明. 将式子左端展开为级数:

    0+f(x)sinxx dx=k=0kπ2(k+1)π2f(x)sinxx dx=n=0nπnπ+π2f(x)sinxx dx+n=1nππ2nπf(x)sinxx dx=n=00π2f(xnπ)sin(xnπ)xnπdx+n=10π2f(x+nπ)sin(x+nπ)x+nπdx=0π2f(x)sinxx dx+n=10π2(1)nf(x)(1x+nπ+1xnπ)sinx dx=0π2f(x)sinx[1x+n=1(1)n2xx2n2π2]dx\begin{aligned}& \int_{0}^{+\infty} f(x) \frac{\sin x}{x} \mathrm{~d} x=\sum_{k=0}^{\infty} \int_{\frac{k \pi}{2}}^{\frac{(k+1) \pi}{2}} f(x) \frac{\sin x}{x} \mathrm{~d} x \\=& \sum_{n=0}^{\infty} \int_{n \pi}^{n \pi+\frac{\pi}{2}} f(x) \frac{\sin x}{x} \mathrm{~d} x+\sum_{n=1}^{\infty} \int_{n \pi-\frac{\pi}{2}}^{n \pi} f(x) \frac{\sin x}{x} \mathrm{~d} x \\=& \sum_{n=0}^{\infty} \int_{0}^{\frac{\pi}{2}} f(x-n \pi) \frac{\sin (x-n \pi)}{x-n \pi} \mathrm{d} x+\sum_{n=1}^{\infty} \int_{0}^{\frac{\pi}{2}} f(x+n \pi) \frac{\sin (x+n \pi)}{x+n \pi} \mathrm{d} x \\=& \int_{0}^{\frac{\pi}{2}} f(x) \frac{\sin x}{x} \mathrm{~d} x+\sum_{n=1}^{\infty} \int_{0}^{\frac{\pi}{2}}(-1)^{n} f(x)\left(\frac{1}{x+n \pi}+\frac{1}{x-n \pi}\right) \sin x \mathrm{~d} x \\=& \int_{0}^{\frac{\pi}{2}} f(x) \sin x\left[\color{green}{\frac{1}{x}+\sum_{n=1}^{\infty}(-1)^{n} \frac{2 x}{x^{2}-n^{2} \pi^{2}}}\right] \mathrm{d} x\end{aligned}\\

    绿色部分正是 1sinx\displaystyle\frac{1}{\sin x} 的级数展开. 这个结论又是怎么得到的捏?
    我们偏题推导一下. 考虑函数

    f(x)=cosαx(π<xπ,0<α<1)f(x)=\cos \alpha x(-\pi<x\leq \pi,\,0<\alpha<1)

    计算其 Fourier 级数:

    a0=1πππcosαxdx=2sinαππa_0=\frac{1}{\pi }\int_{-\pi }^{\pi }\cos \alpha x\text{d}x= \frac{2\sin \alpha \pi }{\pi }

    an=1πππcosαxcosnx dx=12πππ[cos(α+n)x+cos(αn)x]dx=sinαππ2αα2n2(n0)\begin{aligned}a_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} \cos \alpha x \cos n x \mathrm{~d} x =\frac{1}{2 \pi} \int_{-\pi}^{\pi}[\cos (\alpha+n) x+\cos (\alpha-n) x]\mathrm{d} x \\&=\frac{\sin \alpha \pi}{\pi} \cdot \frac{2 \alpha}{\alpha^{2}-n^{2}} \qquad(\forall n \geq 0)\end{aligned}\\

    bn=1πππcosαxsinnx dx=12πππ[sin(α+n)x+sin(αn)x]dx=0b_{n} =\frac{1}{\pi} \int_{-\pi}^{\pi} \cos \alpha x \sin n x \mathrm{~d} x =\frac{1}{2 \pi} \int_{-\pi}^{\pi}[\sin (\alpha+n) x+\sin (\alpha-n) x]\mathrm{d} x =0\\

    故有

    f(x)=sinαπαπ+2αsinαππn=1+(1)nα2n2cosnxf(x) = \frac{\sin \alpha \pi}{\alpha \pi}+\frac{2 \alpha \sin \alpha \pi}{\pi} \sum_{n=1}^{+\infty}\frac{(-1)^{n}}{\alpha^{2}-n^{2}} \cos n x

    因为 f(x)f(x)(π,π)(-\pi, \pi) 上连续且 f(0)=1f(0)=1, 由 Fourier 级数的收敛定理得

    sinαπαπ+2αsinαππn=1+(1)nα2n2cosnx=1\frac{\sin \alpha \pi}{\alpha \pi}+\frac{2 \alpha \sin \alpha \pi}{\pi} \sum_{n=1}^{+\infty}\frac{(-1)^{n}}{\alpha^{2}-n^{2}} \cos n x=1

    α=xπ\displaystyle\alpha=\frac{x}{\pi}, 可得

    sinxx+n=1+(1)n2xsinxx2n2π2=1\frac{\sin x}{x}+\sum_{n=1}^{+\infty} \frac{(-1)^{n} 2 x \sin x}{x^{2}-n^{2} \pi^{2}}=1

    整理即得

    1sinx=1x+n=1(1)n2xx2n2π2\color{Green} {\frac{1}{\sin x}=\frac{1}{x}+\sum_{n=1}^{\infty}(-1)^{n} \frac{2 x}{x^{2}-n^{2} \pi^{2}}}

    故 Lobachevsky 积分法得证.

  • 解法二: 利用 Fourier 变换
    考虑函数

    f(x)={1,x<112,x=10,x>1f(x)=\left\{\begin{matrix}1,\quad |x|<1\\\frac{1}{2},\quad |x|=1\\0,\quad |x|>1\end{matrix}\right.\\

    对其使用 Fourier 变换得

    F(ω)=F[f(t)]=11cosωxdx=2sinωωF(\omega )=\mathcal{F} [f(t)]=\int_{-1}^{1}\cos \omega x\text{d}x= 2\frac{\sin\omega}{\omega }\\

    再求 Fourier 逆变换得

    f(x)=12π2sinωcosωxωdωf(x)=\frac{1}{2\pi }\int_{-\infty }^{\infty }\frac{2\sin\omega \cos \omega x}{\omega }\text{d}\omega

    x=0x=0 即得

    1=12π2sinωωdω1=\frac{1}{2\pi }\int_{-\infty }^{\infty }\frac{2\sin\omega }{\omega }\text{d}\omega

    整理即得 0+sinxxdx=π2\color{red}{\displaystyle\int_{0}^{+\infty}\frac{\sin x}{x}\text{d}x=\frac{\pi}{2}}\\.

  • 解法三: 利用 Laplace 变换:
    熟知 Laplace 变换具有这样的性质:

    sL[f(t)]ds=L[f(t)t]\int_{s}^{\infty }\mathcal{L} [f(t)] \text{d}s=\mathcal{L}\left [ \frac{f(t)}{t} \right ] \\

    利用该性质可以轻松解决 Dirichlet 积分:

    0sinttdt=lims00estsinttdt=lims0L[sintt]=lims0sL[sint]ds=lims0sdss2+1=lims0[π2arctan(s)]=π2\begin{aligned}\color{Red} {\int_{0}^{\infty} \frac{\sin t}{t} \mathrm{d} t} &=\lim _{s \rightarrow 0} \int_{0}^{\infty} e^{-s t} \frac{\sin t}{t}\mathrm{d}t=\lim _{s \rightarrow 0} \mathcal{L}\left[\frac{\sin t}{t}\right]=\lim _{s \rightarrow 0} \int_{s}^{\infty}\mathcal{L}[\sin t]\text{d}s \\&=\lim _{s \rightarrow 0} \int_{s}^{\infty} \frac{\mathrm{d} s}{s^{2}+1}=\lim _{s \rightarrow 0}\left[\frac{\pi}{2}-\arctan (s)\right]=\color{Red}{\frac{\pi}{2}} \end{aligned}\\

另外, 我们介绍该积分有扩展形式

0+sinaxxdx=π2sgn(a)\int_{0}^{+\infty}\frac{\sin ax}{x}\text{d}x=\frac{\pi}{2}\text{sgn}(a)

Part 6. 反正切函数级数部分

我们知道

arctanx+arctany=arctanx+y1xy(xy<1)\arctan x+\arctan y=\arctan\frac{x+y}{1-xy}\qquad (xy<1)\\

于是

\begin{align}\color{red}{\sum_{n = 1}^{\infty }\arctan \frac{2}{n^2} }& = \sum_{n = 1}^{\infty }\arctan\frac{(1+n)+(1-n)}{1-(1+n)(1-n)} = \sum_{n = 1}^{\infty }\left [\arctan(1+n)+\arctan(1-n) \right ] \\&=\lim_{m \to \infty} \sum_{n = 1}^{m}\left [\arctan(1+n)-\arctan(n-1) \right ] \\&=\lim_{m \to \infty}\left [ \arctan(m+1)+\arctan(m)-\arctan1-\arctan0 \right ] =\color{red}{\frac{3\pi }{4} }\end{align}

Part 7. Laplace 积分部分

取极限的积分部分是所谓 Laplace 积分的一个变体. 积分

0+cosbxx2+a2dx(a,b>0)\int_{0}^{+\infty} \frac{\cos bx}{x^2+a^2}\text{d}x\qquad (a,b>0) \\

称为 Laplace 积分.

  • 解法一: 运用 Laplace 变换

    I(b)=0+cosbxx2+a2dxI(b)=\int_{0}^{+\infty} \frac{\cos bx}{x^2+a^2}\text{d}x \\

    对其作 Laplace 变换

    F(s)=L[I(b)]=0+0+cosbxx2+a2esbdxdb=0+esbcosbxdb0+1a2+x2dxF(s)=\mathcal{L} \left [ I(b) \right ] =\int_{0}^{+\infty}\int_{0}^{+\infty} \frac{\cos bx}{x^2+a^2}e^{-sb}\text{d}x\text{d}b =\int_{0}^{+\infty}e^{-sb}\cos bx\text{d}b \int_{0}^{+\infty}\frac{1}{a^2+x^2}\text{d}x

    运用熟知的余弦函数的 Laplace 变换公式

    0+esbcosbxdb=L[cos(bx)]=ss2+x2\int_{0}^{+\infty}e^{-sb}\cos bx\text{d}b =\mathcal{L} \left [ \cos (bx) \right ]= \frac{s}{s^2+x^2}\\

    得到

    F(s)=0+esbcosbxdb0+1a2+x2dx=0+ss2+x21a2+x2dxF(s)=\int_{0}^{+\infty}e^{-sb}\cos bx\text{d}b \int_{0}^{+\infty}\frac{1}{a^2+x^2}\text{d}x =\int_{0}^{+\infty}\frac{s}{s^2+x^2}\frac{1}{a^2+x^2}\text{d}x\\

    右端积分稍作变形:

    0+ss2+x21a2+x2dx=0+s2a2(s2+x2)(a2+x2)ss2a2dx=ss2a20+(1a2+x21s2+x2)dx\int_{0}^{+\infty}\frac{s}{s^2+x^2}\frac{1}{a^2+x^2}\text{d}x=\int_{0}^{+\infty}\frac{s^2-a^2}{\left ( s^2+x^2 \right ) \left ( a^2+x^2 \right )}\frac{s}{s^2-a^2}\text{d}x=\frac{s}{s^2-a^2}\int_{0}^{+\infty}\left ( \frac{1}{a^2+x^2}-\frac{1}{s^2+x^2} \right ) \text{d}x\\

    因为

    (1a2+x21s2+x2)dx=1aarctanxa1sarctanxs+C\int \left ( \frac{1}{a^2+x^2}-\frac{1}{s^2+x^2} \right ) \mathrm{d} x=\frac{1}{a}\arctan \frac{x}{a}-\frac{1}{s}\arctan \frac{x}{s}+C \\

    所以

    F(s)=ss2a20+(1a2+x21s2+x2)dx=ss2a2[π2aπ2s]=π2a1s+aF(s)=\frac{s}{s^2-a^2}\int_{0}^{+\infty}\left ( \frac{1}{a^2+x^2}-\frac{1}{s^2+x^2} \right ) \text{d}x=\frac{s}{s^2-a^2}\left [ \frac{\pi }{2a}- \frac{\pi }{2s} \right ] =\frac{\pi}{2a}\frac{1}{s+a}\\

    再对两边进行 Laplace 逆变换即得

    I(b)=0+cosbxx2+a2dx=π2aeabI(b)=\color{red}{\int_{0}^{+\infty} \frac{\cos bx}{x^2+a^2}\text{d}x =\frac{\pi}{2a}e^{-ab}}\\

    将原题积分改写使其具有拉普拉斯积分的形式即可求解:

    limt0+20202020tcosxx2+t2 dx=limt0+20202020cos(txt)(xt)2+1 d(xt)=limt0++costαα2+1 dα=limt0+πet=π\begin{aligned}\lim _{t \rightarrow 0^{+}} \int_{-2020}^{2020} \frac{t \cos x}{x^{2}+t^{2}} \mathrm{~d} x &=\lim _{t \rightarrow 0^{+}} \int_{-2020}^{2020} \frac{\cos \left(t \frac{x}{t}\right)}{\left(\frac{x}{t}\right)^{2}+1} \mathrm{~d}\left(\frac{x}{t}\right)=\lim _{t \rightarrow 0^{+}} \int_{-\infty}^{+\infty} \frac{\cos t \alpha}{\alpha^{2}+1} \mathrm{~d} \alpha =\lim _{t \rightarrow 0^{+}} \pi e^{-t}=\color{red}\pi\end{aligned}\\

  • 解法二: 解微分方程
    仍然设

    I(b)=0+cosbxx2+a2dxI(b)=\int_{0}^{+\infty} \frac{\cos bx}{x^2+a^2}\text{d}x \\

    对其求导得到

    I(b)=0+xsinbxa2+x2dx=0+(a2+x2a2)sinbxx(a2+x2)dx=0+sinbxxdx+a20+sinbxx(a2+x2)dx=π2+a20+sinbxx(a2+x2)dx\begin{aligned}I^{\prime}(b) &=-\int_{0}^{+\infty} \frac{x \sin b x}{a^{2}+x^{2}} \mathrm{d} x =-\int_{0}^{+\infty} \frac{\left(a^{2}+x^{2}-a^{2}\right) \sin b x}{x\left(a^{2}+x^{2}\right)} \mathrm{d} x \\&=-\int_{0}^{+\infty} \frac{\sin b x}{x} \mathrm{d} x+a^{2} \int_{0}^{+\infty} \frac{\sin b x}{x\left(a^{2}+x^{2}\right)} \mathrm{d} x \\&=-\frac{\pi}{2}+a^{2} \int_{0}^{+\infty} \frac{\sin b x}{x\left(a^{2}+x^{2}\right)} \mathrm{d} x \end{aligned}\\

    I(b)=a20+cosbxa2+x2dx=a2I(b)I^{\prime \prime}(b) =a^{2} \int_{0}^{+\infty} \frac{\cos b x}{a^{2}+x^{2}} d x=a^{2} I(b)\\

    于是得到一个二阶微分方程:

    I(b)a2I(b)=0I^{\prime \prime}(b)-a^{2} I(b)=0\\

    容易得到初值条件

    I(0)=π2a,I(0)=π2I(0)=\frac{\pi}{2a},\quad I^{\prime}(0)=-\frac{\pi}{2}\\

    解微分方程得到

    I(b)=C1eab+C2eabI(b)=C_1e^{ab}+C_2e^{-ab}\\

    由初值条件得到 C1=0,C2=π2C_1=0, C_2=\frac{\pi}{2}. 于是

    I(b)=0+cosbxx2+a2dx=π2aeabI(b)=\color{red}{\int_{0}^{+\infty} \frac{\cos bx}{x^2+a^2}\text{d}x =\frac{\pi}{2a}e^{-ab}}\\

Part 8. 极限部分

分步解 limn{[(01xn11+x dx)n12]n2}\displaystyle\lim _{n \rightarrow \infty}\left\{\left[\left(\int_{0}^{1} \frac{x^{n-1}}{1+x} \mathrm{~d} x\right) n-\frac{1}{2}\right] \frac{n}{2}\right\}.

先将小括弧里的积分计算出来(分部积分法:

01xn11+x dx=1n[xn1+x01+01xn(1+x)2 dx]=1n[12+01xn(1+x)2 dx]\int_{0}^{1} \frac{x^{n-1}}{1+x} \mathrm{~d} x=\frac{1}{n}\left[\left.\frac{x^{n}}{1+x}\right|_{0} ^{1}+\int_{0}^{1} \frac{x^{n}}{(1+x)^{2}} \mathrm{~d} x\right]=\frac{1}{n}\left[\frac{1}{2}+\int_{0}^{1} \frac{x^{n}}{(1+x)^{2}} \mathrm{~d} x\right]\\

所以原极限变为

limn{[(01xn11+x dx)n12]n2}=limn{n201xn(1+x)2 dx}\lim _{n \rightarrow \infty}\left\{\left[\left(\int_{0}^{1} \frac{x^{n-1}}{1+x} \mathrm{~d} x\right) n-\frac{1}{2}\right] \frac{n}{2}\right\}=\lim _{n \rightarrow \infty}\left\{\frac{n}{2} \int_{0}^{1} \frac{x^{n}}{(1+x)^{2}} \mathrm{~d} x\right\}\\

继续分部计算积分

limn{n201xn(1+x)2 dx}=limn{n2(n+1)[xn+1(1+x)201+201xn+1(1+x)3 dx]}=limn{n8(n+1)+nn+101xn+1(1+x)3 dx}=18\begin{aligned}\lim _{n \rightarrow \infty}\left\{\frac{n}{2} \int_{0}^{1} \frac{x^{n}}{(1+x)^{2}} \mathrm{~d} x\right\} &=\lim _{n \rightarrow \infty}\left\{\frac{n}{2(n+1)}\left[\left.\frac{x^{n+1}}{(1+x)^{2}}\right|_{0} ^{1}+2 \int_{0}^{1} \frac{x^{n+1}}{(1+x)^{3}} \mathrm{~d} x\right]\right\} \\&=\lim _{n \rightarrow \infty}\left\{\frac{n}{8(n+1)}+\frac{n}{n+1} \int_{0}^{1} \frac{x^{n+1}}{(1+x)^{3}} \mathrm{~d} x\right\} \\&=\color{red}{\frac{1}{8}}\end{aligned}\\

合并 Part 1-8

将各部分结果代入原题计算即得

0+ess5 ds2++et22 dt0+sint2 dt(n=0(1)n2n+10+sinxx dx+n=1arctan2n2limt0+20202020tcosxx2+t2 dx)limn{[(01xn11+x dx)n12]n2}=520 \frac{\frac{\displaystyle {\int_{0}^{+\infty} e^{-s} s^{5} \mathrm{~d} s}}{2}+\frac{\displaystyle \int_{-\infty}^{+\infty} e^{-\frac{t^{2}}{2}} \mathrm{~d} t}{\displaystyle \int_{0}^{+\infty} \sin t^{2} \mathrm{~d} t}\left(\frac{\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1}}{\displaystyle\int_{0}^{+\infty} \frac{\sin x}{x} \mathrm{~d} x}+\frac{\displaystyle\sum_{n=1}^{\infty} \arctan \frac{2}{n^{2}}}{\displaystyle\lim _{t \rightarrow 0^{+}} \displaystyle \int_{-2020}^{2020} \frac{t \cos x}{x^{2}+t^{2}} \mathrm{~d} x}\right)}{\displaystyle\lim _{n \rightarrow \infty}\left\{\left[\left(\displaystyle\int_{0}^{1} \frac{x^{n-1}}{1+x} \mathrm{~d} x\right) n-\frac{1}{2}\right] \frac{n}{2}\right\}}=520\\